At a school event, adults are seated in groups of exactly 6 and children are seated in groups of exactly 4. If there are the same number of adults as children, what is the minimum number of adults attending?

To solve this problem, let's introduce some variables:

- Let \( A \) represent the total number of adults attending the event.
- Let \( C \) represent the total number of children attending the event.

According to the problem:

1. The adults are seated in groups of exactly 6, so the total number of adults must be a multiple of 6.
2. The children are seated in groups of exactly 4, so the total number of children must be a multiple of 4.
3. The number of adults is equal to the number of children.

Given these points, we can set up the following equation:

\[
A = C
\]

Since \( A \) (the number of adults) must be a multiple of 6, we can write:

\[
A = 6k \quad \text{for some integer } k
\]

Similarly, since \( C \) (the number of children) must be a multiple of 4, we can write:

\[
C = 4m \quad \text{for some integer } m
\]

Given that \( A = C \), we can equate these two expressions:

\[
6k = 4m
\]

To find the minimum number of adults, we need to find the smallest integer values of \( k \) and \( m \) such that both sides are equal. We can rewrite this equation as:

\[
\frac{6k}{4} = m \implies \frac{3k}{2} = m
\]

This equation implies that \( m \) must be an integer, which happens if \( k \) is an even number. Let's take the smallest even number for \( k \), which is \( k = 2 \):

\[
m = \frac{3 \cdot 2}{2} = 3
\]

Now, substitute \( k = 2 \) back into the expression for \( A \):

\[
A = 6k = 6 \cdot 2 = 12
\]

So, the number of adults \( A \) is 12. Since \( A = C \), the number of children is also 12.

Therefore, the minimum number of adults attending is **12**.

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How would the solution change if adults were seated in groups of 5 and children in groups of 3?
2. What is the least common multiple (LCM) of 6 and 4, and why is it relevant to this problem?
3. How can you verify that 12 is indeed the smallest number that meets the conditions of this problem?
4. If the number of adults had to be a multiple of 8 instead of 6, what would the new minimum be?
5. What other real-world situations might require finding the least common multiple?

**Tip:** When solving problems involving multiples, identifying the least common multiple (LCM) of the given numbers can often help find the smallest solution.

Discover how to determine the minimum number of adults and children seated in groups of 6 and 4 respectively. This problem involves finding the least common multiple (LCM) and solving simultaneous equations. Suitable for students in Grades 6-8.

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